Question

Question #adffa

Answer 1

`[-sin(x)tan(x)-cos(x)sec^2(x)]/tan^2(x)`

Explanation:

Assuming your problem is

`d/dx[(cos(x))/(tan(x))]`

Then, we have to use the quotient rule, which states that

`d/dx(x/y)=(color(red)(x')color(blue)y-color(green)xcolor(orange)(y'))/color(brown)(y^2)`

Here, let `x=cos(x)`, `y=tan(x)`

So, we get

`d/dx[(cos(x))/(tan(x))]=(color(red)(-sin(x))color(blue)(tan(x))-color(green)(cos(x))color(orange)(sec^2(x)))/color(brown)(tan^2(x))`

Answer 2

`dy/dx=(tanx(-sinx)-cosxsec^2x)/(tanx)^2`

`"We can simplify further if necessary":`

Explanation:

`"Let "y=cosx/tanx`

`"We need to apply [quotient rule]"`

(https://socratic.org/calculus/basic-differentiation-rules/quotient-rule)

`"Let" \ u=cosx, (du)/dx=-sinx`

`"Let" \ v=tanx, (dv)/dx=sec^2x`

`y=u/v, dy/dx=(v(du)/dx-u(dv)/dx)/v^2`

`"Substituting, we have"`

`dy/dx=(tanx(-sinx)-cosxsec^2x)/(tanx)^2`